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Your first part is correct, i.e. probability of getting 3 tails is 1/8. Similarly, you can compute the probability of getting three heads. Please note that each coin toss is independent of one another, that's why we can multiply the probability of each event. P(HHH) = P(getting head on the first toss) and P(getting head on the second toss) and P(getting head on the third toss)= 1/2 * 1/2 * 1/2 = 1/8

AND implies Multiplication(*) OR implies Addition(+)

Quote:

Because each event is independent from one another, I calculate the probability of getting heads as 1-1/8=7/8 but the right answer is 1/8....

Here you are assuming that following events are complementary to each other: 1. Getting three tails, i.e. P(TTT) and 2. Getting three heads, i.e.P(HHH) P(TTT) + P(HHH) = 1 This is not correct.

Let's enumerate the all possible outcomes(sample space). 1. HHH 2. HHT 3. HTT 4. TTT 5. TTH 6. THH 7. THT 8. HTH

Sum of probabilities of all these events will be equal to 1. i.e. P(HHH) + P(HHT) + P(HTT) + P(TTT) + P(TTH) + P(THH) + P(THT) + P(HTH) = 1

In general, if you toss the coin n times, then total possible outcomes = 2^n. In this cane, n= 3, so total outcomes = 2^3 = 8.

Your first part is correct, i.e. probability of getting 3 tails is 1/8. Similarly, you can compute the probability of getting three heads. Please note that each coin toss is independent of one another, that's why we can multiply the probability of each event. P(HHH) = P(getting head on the first toss) and P(getting head on the second toss) and P(getting head on the third toss)= 1/2 * 1/2 * 1/2 = 1/8

AND implies Multiplication(*) OR implies Addition(+)

Quote:

Because each event is independent from one another, I calculate the probability of getting heads as 1-1/8=7/8 but the right answer is 1/8....

Here you are assuming that following events are complementary to each other: 1. Getting three tails, i.e. P(TTT) and 2. Getting three heads, i.e.P(HHH) P(TTT) + P(HHH) = 1 This is not correct.

Let's enumerate the all possible outcomes(sample space). 1. HHH 2. HHT 3. HTT 4. TTT 5. TTH 6. THH 7. THT 8. HTH

Sum of probabilities of all these events will be equal to 1. i.e. P(HHH) + P(HHT) + P(HTT) + P(TTT) + P(TTH) + P(THH) + P(THT) + P(HTH) = 1

In general, if you toss the coin n times, then total possible outcomes = 2^n. In this cane, n= 3, so total outcomes = 2^3 = 8.

Hope it helps.

Thanks.

Thanks for the explanation, this is very helpful.

Could you help me understand the difference between this and the following problem?

A fair coin is flipped twice. What is the probability that the coin lands showing heads on the first flip, second flip or both?

The solution to this question is two-step: 1) P (two tails)=1/4, 2) 1-1/4=3/4

1. Compute the probability of the favorable event In this case required probability = P(HH) + P(HT) + P(TH) =\(\frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{2} = \frac{3}{4}\) or 2. Compute the probability of complementary event and subtract it from 1. You have solved it using this method.

In general, when a complementary event is small then use the 2nd approach. The complementary approach can also be used in P&C questions.

Please go through the first post of the thread. These points are discussed in detail with examples.

Q: If the probability of a certain event is p, what is the probability of it occurring k times in n-time sequence? (Or in English, what is the probability of getting 3 heads while tossing a coin 8 times?) Solution: All events are independent. So, we can say that:

P′=pk∗(1−p)n−kP′=pk∗(1−p)n−k (1)

But it isn't the right answer. It would be right if we specified exactly each position for events in the sequence. So, we need to take into account that there are more than one outcomes. Let's consider our example with a coin where "H" stands for Heads and "T" stands for Tails: HHHTTTTT and HHTTTTTH are different mutually exclusive outcomes but they both have 3 heads and 5 tails. Therefore, we need to include all combinations of heads and tails. In our general question, probability of occurring event k times in n-time sequence could be expressed as:

P=Cnk∗pk∗(1−p)n−kP=Ckn∗pk∗(1−p)n−k (2)

In the example with a coin, right answer is P=C83∗0.53∗0.55=C83∗0.58

Aren't we looking for the permutations here instead of the combination ? aren't HHHTTTTT and HHTTTTTH the same combination but two different permutations? Thank you for your clarifications

Aren't we looking for the permutations here instead of the combination? aren't HHHTTTTT and HHTTTTTH the same combination but two different permutations?

I have a question regarding solving the below example using reverse combinatorial method:

*********************** Example #2 Q: Given that there are 5 married couples. If we select only 3 people out of the 10, what is the probability that none of them are married to each other? Solution: 2) reversal combinatorial approach: In this example reversal approach is a bit shorter and faster. C51C15 - we choose 1 couple out of 5 couples. C81C18 - we chose one person out of remaining 8 people. C103C310 - the total number of combinations to choose 3 people out of 10 people. p=1−C51∗C81C103=1−5∗810∗3∗4=23 ************************************** My logic is as below: We need to choose 3 people, 1 couple and 1 other person The first person can be anyone..so 10 possibilities second has to be the partner of the first..so only 1 possibility third can be any of the remaining 8..so 8 possibility so total=10 x 1 x 8 now arrangement should not matter, so i have to divide by 3! So total unfavorable options=10 x 8 /3!=40/3

But according to the official answer, total unfavorable options =8 x 5=40 I understand the logic of the official answer, I just can't figure out where I am going wrong with my logic...

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